If y is a function of x and log(x+y) = 2xy,th | vedclass

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(A) Given the equation: $\log(x+y) = 2xy$ ... $(i)$ At $x = 0$,substituting into $(i)$ gives: $\log(0+y) = 2(0)y \implies \log(y) = 0 \implies y = e^0

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$1$

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(A) Given the equation: $\log(x+y) = 2xy$ ... $(i)$ At $x = 0$,substituting into $(i)$ gives: $\log(0+y) = 2(0)y \implies \log(y) = 0 \implies y = e^0 = 1$. Now,differentiate both sides of $(i)$ with respect to $x$: $\frac{d}{dx}(\log(x+y)) = \frac{d}{dx}(2xy)$ $\frac{1}{x+y} \cdot (1 + y') = 2(y + xy')$ Substitute $x = 0$ and $y = 1$ into the differentiated equation: $\frac{1}{0+1} \cdot (1 + y'(0)) = 2(1 + 0 \cdot y'(0))$ $1 \cdot (1 + y'(0)) = 2(1)$ $1 + y'(0) = 2$ $y'(0) = 2 - 1 = 1$ Thus,the value of $y'(0)$ is $1$.

If $y$ is a function of $x$ and $\log(x+y) = 2xy$,then the value of $y'(0)$ is

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